Why can't the limit of $A(n)$ as $n$ tends to $\infty$ be considered the supremum here?

Question

I have the following exercise

$A=\{\frac{n+1}{n+2}, n\in \mathbb{N}\}$

We have to find its supremum

We did that in class and what happened was that we showed $1$ is an upper bound of $A$ as $\lim_{n\rightarrow \infty} A=1$ then we showed there are not any upper bounds less than $1$ by assuming the opposite and reaching a contradiction, I was wondering why is the second step needed in the first place? I talked about it with my professor and I wasn't convinced as he ended up saying something along the lines of "the supremum doesn't have to be in the set"

The reason I thought this way is that how I understand this limit is "A gets infinitely close to $1$ as $n$ tends to $\infty$ but never really touches it " so I think that the lowest upper bound possible as there is no number $x$ such that $A(n)\leq x<1$ if it gets infinitely close.

Answer 1

First of all, a warning in your notation. The way it is written, $A$ is a set. It is a set defined by a sequence, but nevertheless, it is a set. This means that writing

$$\lim_{n\to\infty} A$$

is, mathematically speaking, complete nonsense.

If you have a sequence $(x_n)_{n=1}^\infty$, and you use it to define a set $X$ as $X=\{x_n|n\in\mathbb n\}$, then you can talk about the limit of the sequence, but you cannot talk about the limit of the set.

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With that out of the way, your main question is "why is the second step needed in the first place", and the answer is simple:

The second step is needed because $2$ is also an upper bound of $A$, but it is not a supremum. It's as simple as that.

Now, there are several alternative ways you could whow that the supremum of $A$ is $1$, but just saying "well the limit of $A$ is $1$, so..." will simply not cut it. For example, take the sequence

$$\frac{(-1)^n}{n}$$

The limit of this sequence is $0$, however, the supremum of $$\left\{\left.\frac{(-1)^n}{n}\right|n\in\mathbb N\right\}$$ is actually $\frac12$.

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Note:

for practice you could try and prove the following statement:

If $(x_n)_{n\in\mathbb N}$ is an increasing sequence of real numbers, then the set $X=\{x_n|n\in\mathbb N\}$ has a supremum if and only if the sequence converges. If the sequence converges, then the limit of the sequence is equal to the supremum of the set.

If you prove this statement, then you can use the statement to just say "the limit is the supremum" (so long as you prove that your sequence is increasing, which is simple because its $n$-th element is

$$\frac{n+1}{n+2}=\frac{n+2-1}{n+2}=1-\frac{1}{n+2}$$